Let $G$ be a finite group and let $\mathbb{Q}G=M_{n_1}(D_1)\times\cdots\times M_{n_k}(D_k)$ be the decomposition of $\mathbb{Q}G$ as a product of rings of matrices over divisions rings. Let $Z_i$ be the center of $D_i$. Then each $Z_i$ is an algebraic extension of $\mathbb{Q}$. The question is: how much is it known about these algebraic number fields?. More precisely, for which family of groups $G$ it is known that each $Z_i$ is a Galois extension of $\mathbb{Q}$ and in such a case what can we say about the corresponding Galois groups?.

In more concrete terms (without mention of Wedderburn's theorem), you're asking about the structure of the center of $\mathbf{Q}[G]$ (as a finite product of number fields).
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user30379Jan 23 '13 at 15:24

2 Answers
2

The short answer is yes, the centres $Z_i$ are contained in the cyclotomic extension $E_n= {\mathbb Q}(e^{(2\pi i)/n})$ where $n$ is the order of the group. As pranavk says, we need only prove that the simple factors of the centre of ${\mathbb Q}[G]$ are contained on $E_n$. But the centre is spanned by the averages $ C_x =\sum gxg^{-1}$ where the sum is over all the elements of $G$. Given an absolutely irreducible representation $V$ of $G$ of dimension $r$ , each $C_x$ acts by a scalar $\lambda $ say. By taking traces, you see that $\lambda $ is a rational multiple of the trace of $x$; since $x$ has order dividing $n$, all its eigenvalues in $V$ are $n$-th roots of unity, and hence the trace of $x$ lies in $E_n$.

Now,in your notation, each $M_{n_i}(D_i)$ after tensoring (over $Z_i$) with the algebraic closure of ${\mathbb Q}$ is of the form $End (V)$ for some absolutely irreducible representation $V$ of $G$, hence each $Z_i$ (which as pranavk observed, lies in the centre of ${\mathbb Q}[G]$) lies in $E_n$.

I believe all this is worked out in Serre's book on finite groups (Springer notes) and is well known to experts (I am not one!).

@Aakumadula: Just to be precise, in the 2nd paragraph you're tensoring against $\overline{\mathbf{Q}}$ over $Z_i$ (using some embedding) rather than over $\mathbf{Q}$. Also, though it is an irrelevant refinement, the argument shows that we can take $n$ to be the exponent of $G$ and not just the order (as is also mentioned in Serre's book, if I remember correctly).
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user30379Jan 23 '13 at 16:03

I would like to add that sometimes, the fields could be smaller. For example, if $G=S_n$, then the theory of Young diagrams shows that all the fields $Z_i$ are ${\mathbb Q}$.
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VenkataramanaJan 23 '13 at 17:03

The simplest case of your question is the case where $G$ is the cyclic group of order $n$, it is known
that $\mathbb{Q}G\simeq\mathbb{Q}[X]/(X^n-1)$.
As $X^n-1=\Pi_{d|n}\Phi_d(X)$ where $\Phi_d(X)$ is the $d$-th
cyclotomic polynomial, it follows that
$\mathbb{Q}G\simeq\Pi_{d|n}\mathbb{Q}(\zeta_d)$ where $\zeta_d\in\mathbb{C}$ is a
primitive $d$-th root of unity.

Edit: This is also a very particular but concrete case of the idea of
@Aakumadula.